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Vector calculus, + finding parametric equation

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data
    The line L
    L1: x=3+2t, y=2t, z=t
    Intersects the plane x+3y-z=-4 at a point P. Find a set of parametric equations for the line in the same plane that goes through P and is perpendicular to L.

    2. Relevant equations
    cross-product
    r=r0+t(vector) this is to get in parametric form typically

    3. The attempt at a solution
    Well lets just say first I'm having trouble visually what this question is asking (is there a program where I can graph this type of stuff in 3 space?)

    substituting the values of "L" into the plane equation and then solved for "t" which was -1. I then plugged those values into the parametric equations of the line to get the point of intersection P(1,2,-1)

    Here is where I get confused and visually a bit shaky. I decided to use the normal direction vector, I'll call it
    "n1" from the plane=<1,3,-1> and "n2" from the line =<2,2,1>
    I take the cross product with the resultant vector being <-5,3,4>
    So I then use the parametric equation which is: L2 => x=1-5t, y=-2+3t, z=-1+4t

    I'm think I'm done at this point because now I have to sets of parametric equations. To check my answer I set
    L1 and L2 equal to each other, solved for the variables using elimination and found the intersection point to be the same as I found earlier.

    I have no idea if I'm right or wrong, any tips would be helpful! Please don't just tell me the answer, I would prefr someone to point out a mistake and let me figure out the rest, after all, I'm taking math to actually grasp it :)

    Thanks in advance!
    Cheers
     
  2. jcsd
  3. Feb 10, 2015 #2

    Dick

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    Science Advisor
    Homework Helper

    The intersection point is <1,-2,-1> but that's just a typo. Looks ok otherwise.
     
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